A 5.0-L air sample containing H2S at STP is treated with a catalyst to promote the reaction, H2S + O2 = H2O + S(solid). If 3.2 g of solid S was collected, what is the volume percentage of H2S in the original sample?

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A 5.0-L air sample containing H2S at STP is treated with a catalyst to promote the reaction, H2S + O2 = H2O + S(solid). If 3.2 g of solid S was collected, what is the volume percentage of H2S in the original sample?

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Answer 1 · 2014-08-31T15:13:32

The concentration H₂S is 45 % (v/v).

Step 1. Write the balanced chemical equation.

2H₂S + O₂ → 2H₂O + 2S

Step 2. Convert grams of S → moles of S → moles of H₂S

3.2 g S × $\frac{1 mol S}{32.06 g S} \times \frac{2 {mol H}_{2} S}{2 mol S}$ $(1"mol S")/(32.06"g S") × (2"mol H"_2"S")/(2"mol S")$ = 0.0998 mol H₂S (2 significant figures + 1 guard digit)

Step 3. Use the Ideal Gas Law to calculate the volume of H₂S at STP.

$P V = n R T$ $PV = nRT$

$V = \frac{n R T}{P} = \frac{0.0998 mol \times 8.314 {kPa\cdotL\cdotK}^{- 1} {mol}^{- 1} \times 273.15 K}{100 kPa}$ $V = (nRT)/P = (0.0998"mol" × 8.314 "kPa·L·K"^-1"mol"^-1 × 273.15"K")/(100"kPa")$ = 2.27 L

Step 3. Calculate the concentration of the H₂S.

Percent by volume = $\frac{{volume of H}_{2} S}{volume of air}$ $("volume of H"_2"S")/"volume of air"$ × 100 % = $\frac{2.27 L}{5.0 L}$ $(2.27"L")/(5.0"L")$ × 100 % = 45 % (v/v)

Note: The definition of Standard Pressure was changed to 100 kPa in 1982.

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A 5.0-L air sample containing H2S at STP is treated with a catalyst to promote the reaction, H2S + O2 = H2O + S(solid). If 3.2 g of solid S was collected, what is the volume percentage of H2S in the original sample?

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